Multiply the following rational expressions and simplify the result. $\dfrac{3n^4-12n^2x^2}{8nx^2-4n^2x} \cdot \dfrac{-8x}{n^2-2nx-8x^2}=$
Explanation: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $3n^4-12n^2x^2$, of the first expression can be factored as $3n^2(n+2x)(n-2x)$ by factoring out $3n^2$ and using the difference of squares pattern. The denominator, $8nx^2-4n^2x$, of the first expression can be factored as $4nx(2x-n)$ by factoring out $4nx$. The numerator, $-8x$, of the second expression cannot be factored further. The denominator, $n^2-2nx-8x^2$, of the second expression can be factored as $(n+2x)(n-4x)$ using the sum-product pattern. Now the product looks as follows: $\dfrac{3n^2(n+2x)(n-2x)}{4nx(2x-n)} \cdot \dfrac{-8x}{(n+2x)(n-4x)}$ To multiply two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=} \dfrac{3n^2(n+2x)(n-2x)}{4nx(2x-n)} \cdot \dfrac{-8x}{(n+2x)(n-4x)} $ $\begin{aligned} &\!= \dfrac{3n^2(n+2x)(n-2x) \cdot -8x}{4nx(2x-n) \cdot (n+2x)(n-4x)} &\text{Multiply across.}\\\\\\\\ &\!= \! \dfrac{3{\cancel{n}}\cdot n{\cancel{(n\!+\!2x)}}\!{\cancel{(n\!-\!2x) (-1)}}\!2\!\cdot\!{\cancel{4}}\!{\cancel{x}}}{{\cancel{4}}{\cancel{n}}{\cancel{x}}{\cancel{(2x\!-\!n)}}{\cancel{ (n\!+\!2x)}}(n\!-\!4x)} &\text{Cancel out common factors.}\\\\\\\\ &\!=\dfrac{6n}{n-4x} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{6n}{n-4x}$.